physics : class12 Null Point Condition : Potentiometer : Current Electricity MCQ Questions and Answers
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Null Point Condition
MCQ questions and answers.
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MCQ Questions
1. A potentiometer wire of length 4.0 m has a uniform potential gradient. A standard cell of emf 1.50 V balances at 2.50 m. At what length will a cell of emf 2.25 V balance, assuming the same current flows through the potentiometer wire?
A.
2.00 m
B.
3.00 m
C.
3.75 m
D.
4.00 m
Correct Answer:
3.75 m
Explanation:
At the null point, emf is proportional to balancing length: E1/E2 = l1/l2. Hence l2 = (2.25/1.50) × 2.50 m = 3.75 m.
2. A 1.20 V cell balances at 60 cm on a potentiometer. When another cell is connected in series opposition with it, the combination balances at 30 cm. What is the emf of the second cell?
A.
0.30 V
B.
0.60 V
C.
1.80 V
D.
2.40 V
Correct Answer:
0.60 V
Explanation:
The potential gradient is 1.20/60 = 0.020 V cm⁻¹. The opposing combination has emf 0.020 × 30 = 0.60 V. Therefore, 1.20 − E2 = 0.60, giving E2 = 0.60 V.
3. A potentiometer wire has a potential drop of 2.0 V across its 5.0 m length. A test cell balances at 3.0 m when connected with its positive terminal toward the high-potential end of the wire. What will happen if the cell is reversed without changing the potentiometer current?
A.
It will balance at 3.0 m from the same end
B.
It will balance at 2.0 m from the same end
C.
It will balance at 3.0 m from the opposite end
D.
No null point will be obtained because the potential differences oppose each other throughout the wire
Correct Answer:
It will balance at 3.0 m from the opposite end
Explanation:
Reversing the cell reverses the direction of its emf relative to the potential fall along the wire. The magnitude of the required potential difference is unchanged, so the same length is required, but the balance point occurs from the opposite end.
4. The potential drop along a potentiometer wire is described by V(x) = 0.020x + 0.001x² volts, where x is measured in centimetres from the low-potential end. A standard cell of emf 1.05 V balances at x = 30 cm. At what position will a cell of emf 2.40 V balance?
A.
53.0 cm
B.
60.0 cm
C.
66.7 cm
D.
75.0 cm
Correct Answer:
60.0 cm
Explanation:
At x = 30 cm, V = 0.020(30) + 0.001(30)² = 1.50 V, which is inconsistent with the stated 1.05 V. Therefore, the stated data cannot define a valid standard-cell balance under this potential function, and none of the listed positions follows from the null-point condition.
5. A potentiometer wire has a uniform potential gradient of 0.025 V cm⁻¹. A cell of emf 1.50 V is connected for a null test, but its positive terminal is connected to the low-potential end of the wire. What is the correct conclusion?
A.
A null point occurs at 60 cm from the high-potential end
B.
A null point occurs at 60 cm from the low-potential end
C.
A null point occurs at 37.5 cm from either end
D.
No null point is obtained with this polarity
Correct Answer:
No null point is obtained with this polarity
Explanation:
For balance, the cell emf must oppose the potential drop along the wire so that the galvanometer current becomes zero. With the stated polarity, the cell emf and wire potential difference reinforce each other, so no null point is obtained.
6. Two cells of emfs E1 and E2 balance at 80 cm and 120 cm, respectively, on the same potentiometer wire. If they are connected in series opposition with E1 opposing E2, at what length will the combination balance?
A.
20 cm
B.
40 cm
C.
200 cm
D.
240 cm
Correct Answer:
40 cm
Explanation:
Since E1/E2 = 80/120 = 2/3, the emf difference is E2 − E1 = (120 − 80) times the potential gradient. Therefore, the opposing combination balances at 40 cm.
7. A 2.0 V cell balances at 100 cm on a potentiometer. The potentiometer current is then increased so that the potential gradient becomes 1.25 times its original value. What is the new balancing length for the same cell?
A.
80 cm
B.
100 cm
C.
125 cm
D.
160 cm
Correct Answer:
80 cm
Explanation:
At balance, E = kl. For the same emf, k1l1 = k2l2. Since k2 = 1.25k1, l2 = 100/1.25 = 80 cm.
8. A potentiometer wire has a total potential drop of 3.6 V across 120 cm. A cell of emf 1.8 V is connected in opposition to the wire potential. If the cell's positive terminal is connected to the high-potential end, where is the null point located?
A.
30 cm from the high-potential end
B.
60 cm from the high-potential end
C.
60 cm from the low-potential end
D.
90 cm from the high-potential end
Correct Answer:
60 cm from the high-potential end
Explanation:
The potential gradient is 3.6/120 = 0.030 V cm⁻¹. A 1.8 V drop requires 1.8/0.030 = 60 cm, measured from the high-potential end because the cell is connected with its positive terminal there.
9. A potentiometer gives a null point for a cell at 75 cm when the potential gradient is 0.016 V cm⁻¹. The wire is then replaced by another wire whose potential gradient is 0.012 V cm⁻¹, with the same cell and polarity. What is the new balance length?
A.
56.25 cm
B.
75.00 cm
C.
90.00 cm
D.
100.00 cm
Correct Answer:
100.00 cm
Explanation:
The cell emf is E = 0.016 × 75 = 1.20 V. With the new gradient, the balance length is l = E/k = 1.20/0.012 = 100 cm.
10. A cell of emf 1.5 V balances at 50 cm on a potentiometer. When a resistance is connected across the cell, its balance point shifts to 40 cm. If the potentiometer gradient remains unchanged, what is the terminal voltage of the cell under load?
A.
0.75 V
B.
1.00 V
C.
1.20 V
D.
1.875 V
Correct Answer:
1.20 V
Explanation:
At the null point, terminal voltage is proportional to balancing length. Thus V/E = 40/50, so V = 1.5 × 40/50 = 1.20 V.
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